find the equation of circle passing through
(0, 1), (2, -1) and (32)
Answers
Answer:
2x² + 2y² - 7x - 3y - 13 = 0
Step-by-step explanation:
Let the circle be x² + y² + Dx + Ey + F = 0
P(0, 1) : 0 + 1² + E + F = 0
P(2, -1) : 2² + 1² + 2D - E + F = 0
P(3, 2) : 3² + 2² + 3D + 2E + F = 0
E + F = -1 ….. (1)
2D - E + F = -5 ………(2)
3D + 2E + F = -13 …….(3)
Simplification
(1) + (2), 2D + 2F = -6 ==> D + F = -3 ……. (4)
(3) - (2), D + 3E = -8 ==> D = -3E - 8 ….. (5)
(3) - (1), 3D + E = -12 …………(6)
Substitute (5) in (6), 3(-3E - 8) + E = -12
==> -9E - 24 + E = -12
==> 8E = -12
==> E = -3/2
(5) ==> D = -3E - 8 = 9/2 - 8 = - 7/2
==> D = -7/2
(4) ==> D + F = -3
==> F = -3 - 7/2
==> F = -13/2
The equation of the circle is
x² + y² +(-7/2)x +(-3/2)y +(-13/2) = 0
==> x² + y² - 7/2 x - 3/2 y - 13/2 = 0
==> 2x² + 2y² - 7x - 3y - 13 = 0
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